Try putting this through your GHCI:
:t twice f x = f (f x)
I'd presume that based on the inference of (f x) f is (t -> t) and x :: t
Yes, Maybe I should get the right associativity rule cleared first.
Cheers,
Paul
At 20:35 01/04/2008, you wrote:
>PR Stanley:
>>I know sort of instinctively why it is so but can't think of the
>>formal rationale for it:
>>f g x = g (g x) :: (t -> t) -> (t -> t)
>>First of all - it is not the definition f g x = ... :: (t-> ...
>but the type of the function which might be specified:
>f :: (t->t)->t->t
>Then, the answer to:
>>Why not
>>(t -> t) -> t -> (t -> t)
>>to take account of the argument x for g?
>>is simple. If t is the type of x, then g must be g :: t->t, you're right.
>So f :: (t->t) -> t -> [the type of the result]
>But this result is of the type t, it is g(g x), not (t->t), it is as
>simple as that. Perhaps you didn't recognize that "->" is syntactically
>a right-associative op, so
>a->b->c is equivalent to a->(b->c), or
>(t->t)->t->t equiv. to (t->t)->(t->t)
>>Jerzy Karczmarczuk
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